cam dynamics-02 tiberiu-petrescu

7/31/2019 Cam Dynamics-02 Tiberiu-petrescu

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Al III-lea Seminar National de Mecanisme, Craiova, 2008 1

DYNAMICS OF CAM GEARS ILLUSTRATED AT THECLASSIC DISTRIBUTION MECHANISM

Relly Victoria PETRESCU, Adriana COM NESCU, Florian IonPETRESCU

Abstract: The paper presents an original method in determining the general dynamicof the mechanisms with rotation cams and followers, particularized to the platetranslated follower. First, one presents the dynamics kinematics. Then one solves the

Lagrange equation and with an original dynamic model (see [1]) with one degree of freedom, with variable internal amortization, one makes the dynamic analyze of twomodels.

1 Introduction

The paper proposes an original dynamic model illustrated for therotating cam with plate translate follower. One presents the dynamicskinematics (the original kinematics); the variable velocity of the camshaftobtained by an approximately method has been used with an original dynamicsystem having one grade of mobility and an variable internal amortization [1];one tests two movement laws, one classics and one original.

2 Dynamics of the classic distribution mechanism

2.1 Precision kinematics at the classic distribution mechanism

In the picture number one, one presents the kinematics schema of the classicdistribution mechanism, in two consecutive positions; with a interrupted line isrepresented the particular position when the follower is situated in the mostdown plane, (s=0), and the cam which has an orally rotation, with constantangular velocity, , one is situated in the point A 0, (the recordation point

between the base profile and the up profile), particular point that mark the up begin of the follower, imposed by the cam-profile; with a continue line isrepresented the superior couple in someone position of the up phase.


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Al III-lea Seminar National de Mecanisme, Craiova, 20082

O

A i

r 0 =s 0

s

s

r A

A 0

1vr

2vr

12vr

B

C

D

A 0i

&

Fig. 1. The kinematics of the classic distribution mechanism

The point A 0, which marks the initial couple position, represents in thesame time the contact point between the cam and follower in the first position.The cam has an angular velocity, , (the camshaft angular velocity).

Cam is rotating with the velocity, , describing the angle , which showhow the base circle has rotated in the orally sense, (with the camshaft together);this rotation can be seen on the base circle between the two particular points, A 0 and A 0i.

In this time the vector r A=OA (which represents the distance betweenthe centre of cam O, and the contact point A), has rotating (trigonometric) with

the angle, . If one measures the angle, , which position the general vector, r A in function of the particular vector, r A0, one obtains the relation (0): += (0)

Where r A is the module of the vector, Ar r

, and A represents the phaseangle of the vector, Ar

r.

The rotating velocity of the vector Ar r

is A & which its a function of the

angular velocity of the camshaft, , and a rotating angle, , (by the movementlaws s( ), s(), s( )).


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The follower isnt acted directly by the cam, by the angle, , and by theangular velocity ; its acted by the vector Ar

r, which has the module r A, the

position angle A and the angular velocity A & . From here result a particular

(dynamic) kinematics, the classical kinematics being just a static andapproximate kinematics.

Kinematics one defines the next velocities (see the picture 1).1v

r=the cam velocity; which is the velocity of the vector, Ar

r, in the point

A; now the classical relation (1) become an approximately relation, and the realrelation take the form (2).

.1 Ar v = (1)

A Ar v &.1 = (2)

The velocity AC v ==1r

is separating in the velocity 2vr

=BC (thefollower velocity which act in its axe, on a vertical direction) and 12v

r=AB (the

slide velocity between the two profiles, the sliding velocity between the camand the follower, which works by the direction of the commune tangent line of the two profiles in the contact point).

Usually the cam profile is synthesis with the AD=s knew, for theclassical module C, and one can write the relations:

220

2 ')( ssr r A ++= (3)

220 ')( ssr r A ++= (4)

220

00

')(cos

ssr

sr

r

sr

A ++

+=

+= (5)

220 ')(

''sin

ssr

sr s

r AD

A A ++=== (6)

A A

A A sr s

r vv && '.'

..sin.12 === (7)

Now, the follower velocity isnt s& ( '2 ssv & ), but its given by the

relation (9). At the classical distribution mechanism the transmitting function D,is given by the relations (8):

sv

D

D

A

A

&

&

&

2

.

==

=

(8)

Ds Dssv A === && ''2 (9)

Determining of the sliding velocity between the profiles is made with therelation (10):


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A A

A A sr r sr r vv && ).(..cos. 0

0112 +=

+== (10)

The angles and A will be determined, with him first and secondderivatives.

The angle has been determined from the triangle ODA i (see fig.1) withthe relations (11-13):

220 ')(

'sin

ssr

s

++= (11)

220

0

')(cos ssr

sr

++

+

= (12)

sr s

tg+

=0

' (13)

One derives (11) in function of angle and one obtains (14):

220

0

')(

'''.').('.'.'

cos'.ssr

r

ssssr sr s

A A

++

++

= (14)

The relation (14) will be written in the form (15):

220

220

20

2220

')(].')[(

''.').('''.')'.('cos'.

ssr ssr

sssr ssssr s

++++

+++= (15)

From the relation (12) one extracts the value of cos , which will beintroduced in the left term of the expression (15); then one reduces s.s 2 fromthe right term of the expression (15) and one obtains the relation (16):

220

220

200

220

0

')(].')[(

]')'.(').[(

')('.

ssr ssr

ssr ssr

ssr

sr

++++

++=

++

+ (16)

After some simplifications one obtains finally the relation (17) whichrepresents the expression of :

220

20

')(

')'.(''

ssr

ssr s

++

+= (17)

Now, when one has, explicitly, one can determine the nextderivatives. The expression (17) will be derived directly and one obtains for

begin the relation (18):


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2220

02

022

00

]')[(]'''')][(')(''[2]')][('''2''')('''[

''

ssr

ssssr ssr sssr sssssr s

+++++++++

=

(18)One reduces the terms from the first bracket of the numerator (s.s),

and then one draws out s from the fourth bracket of the numerator and oneobtains the expression (19):

2220

02

022

00

]')[(

]''].[')'.(''.[.2]')].[('''.)'.(''[''

ssr

ssr ssr ssssr sssr s

++

++++++=

(19) Now one can calculate A, with its first two derivatives, A

& i A && . Onewrites, , and not A, to simplify the notation. Now one can determine (20), or (0):

+= (20)One derives (20) and one obtains the relation (21):

.)'1.('. D=+=+=+= &&& (21)One makes the second derivative of (20), and the first derivative of (21)

and one obtains (22):22 ''' ==+= D&&&&&& (22)

One can write now the transmitting functions, D and D (at the classicalmodule, C), in the forms (23-24):

1'+= D (23)''' = D (24)

The follower velocity (25), need the expression of the transmittedfunction, D.

Ds Dssswsv A ===== &&& ''''2 (25)

Where, . Dw = (26)

For the classical distribution mechanism (Module C), the variable w is

the same with A & (see the relation 25). But at the B and F modules (at the camgears where the follower has roll), the transmitted function D, and w, take somecomplex forms.

Now one can determine the acceleration of the follower (27).2

2 )''''( += Ds Dsa y&& (27)In the picture 2, one can see the kinematics classic and dynamic; the

velocities (a), and the accelerations (b).


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a) b)Fig. 2. The classical and dynamic kinematics; a-velocities and b-accelerations

of the follower

To determine the acceleration of the follower, are necessary s and s, Dand D, and .

The kinematics dynamic diagrams of v 2 (obtained with relation 25, seefig. 2, a), and a 2 (obtained with relation 27, see fig. 2, b), have a dynamic aspect

more than one kinematics. One has used the movement law SIN, a rotationvelocity at the crankshaft, n=5500 [rpm], an up angle, u=75 [grad], a downangle d=75 [grad] (identically with the up angle), a ray at the basic circle of thecam, r 0=17 [mm] and a maxim stroke of the follower, h T=6[mm].

Anyway, the dynamic is more complex, having in view the masses andthe inertia moments, the resistant and motor forces, the elasticity constants andthe amortization coefficient of the kinematics chain, the inertia forces of thesystem, the rotation velocity of the camshaft and the variation of the camshaftvelocity, with the cam position, , and with the rotation speed of thecrankshaft, n.

2.2 Solving approximately the Lagrange movement equation

In the kinematics and the static forces study of the mechanisms oneconsiders the shaft velocity constant, =& =constant, and the angular speednull, 0=== &&& . In reality, this velocity it isnt constant, but it is variablewith the camshaft position, .

-4

-3

-2

-1

0

1

2

3

4

0 50 100 150 200

Vclasic[m/s]Vprecis[m/s]

-4000-3000-2000-1000

010002000300040005000

0 50 100 150 200

a2clasic[m/s2]a2precis[m/s2]


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The mechanisms with cam and follower have the variable velocity of the camshaft as well. Then, one shall see the Lagrange equation, written in thedifferentiate mode and its general solve.

The differentiate Lagrange equation, has the form (28):

*2** ..21

. M J J I =+ &&& (28)

Where J* is the inertia moment (mass moment, or mechanic moment) of themechanism, reduced at the crank, and M* represents the difference between the

motor moment reduced at the crank and the resistant moment reduced at thecrank; the angle represents the rotation angle of the crank (crankshaft). J* I represents the derivative of the mechanic moment in function of the rotationangle of the crank (29).

Ld dJ

J I ==

** .

21

.21

(29)

If one use the notation (29), the equation (28) will be written in the form(30):

*2* .. M L J =+ &&& (30)

One divide the terms at J* and (30) takes the form (31):

*

*2

*.

J

M

J

L=+ &&& (31)

The term with 2 & will be move in the right and one obtains (32):

2**

*

. &&& J

L

J

M = (32)

One change the left term of the expression (32) with (33), and one

obtains the relation (34):

...d d

d d

dt d

d d

dt d

==== &&&&

&& (33)

*

2*2

**

* ...

J L M

J L

J M

d d

== (34)

Because, for an angle , vary from the nominal constant value n tothe variable value , one can write the relation (35), where d represents the


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momentary variation for an angle ; the variable d with the constant n driveus to the needed variable, :

d n += (35)

In the relation (35), and d are functions of the angle, , and n is aconstant parameter, which can take different values in function of the rotationvelocity of the drive-shaft, n. To a moment, n is a constant and n is a constantas well (because n is a function of n). The rotation velocity, become afunction of n too (see the relation 36):

))(,()(),( nd nn nn += (36)

With (35) in (34), one obtains the equation (37):

d d J

L

J

M d d nn ].).([).(

2**

*

+=+ (37)

The relation (37) takes the form (38):

]..2)(.[..)(. 22**

*2 d d d

J

Ld

J

M d d nnn ++=+ (38)

The equation (38) will be written in the form (39):

0....2).(.

...)(.

*2

*

2**

*2

=++

+++

d d J L

d d J L

d J L

d J

M d d

n

nn

(39)

The relation (39), take the form (40):

0)...(

.).21

..(2)).(1.(

2**

*

*2

*

=

+++

n

n

d J

Ld

J

M

d d J L

d d J L

(40)

The relation (40) is a equation of the second degree in (d ). Thediscriminate (discriminative) of the equation (40) can be written in the forms(41) and (42):

2*

222*

2

*

*

22*

*2

*

222

2*

2

...).(.

).(.

..4

.).(

nn

nn

n

d J

Ld

J

Ld

J

M

d J

M Ld

J

Ld

J

L

+

+++=(41)


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d J

M d J M Ln .).(.

4 **2

2*

*2++= (42)

One keeps for d , just the positive solution, which can generate positives and negatives normal values (43), and in this mode one obtains for only normal values; for 0


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20

22**20

22

)()(2

)'(])([)(2)2()1(

k K k K

xk s

Dsmk K mk K

K sk K xk sK k k X

T S

++

+

+++

++++=

(48)

20

022

20

22**2

)()(2

)(2)2(

)()(2

)'(])([

k K k K

xk s

sk K xk sK k k

k K k K

xk s

Dsmk K mk K

K

s X T S

+++

+++

++

+

+++=

(49)

2.4 The dynamic analysis

The dynamic analysis or the classical movement law sin, can be seen in thediagram from the picture 3, and in the picture 4 one can see the diagram of an originalmovement law (C4P) (module C).

-2000

-1000

0

1000

2000

3000

4000

5000

6000

0 50 100 150 200

a[m/s2]673.05s*k[mm] k=

n=5000[rot/min] u=75 [grad]k=20 [N/mm]r0=14 [mm]

x0=40 [mm]

hs =6 [mm]

hT=6 [mm]i=1;=8.9%

legea: sin-0y=x-sin(2 x)/(2 )

Analiza dinamic la cama rotativ cu tachettranslant plat - A10

a max =4900

s max =5.78

amin= -1400

Fig. 3. The dynamic analysis of the law sin, Module C, u=75 [grad],

n=5000 [r/m]


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-10000

0

10000

20000

30000

40000

50000

0 20 40 60 80 100 a[m/s2]7531.65s*k[mm] k=

n=10000[rot/min] u=45 [grad]

k=200 [N/mm]r0=17 [mm]

x0=50 [mm]

hs =6 [mm]

hT=6 [mm]i=1; =15.7%

legea:C4P1-1y=2x-x 2

yc=1-x2

Analiza dinamic la cama rotativ cu tachettranslant plat - A10

amax =39000

s max =4.10

amin= -8000

Fig. 4. The dynamic analysis of the new law, C4P, Module C, u=45[grad], n=10000 [r/m]

-40000

-200000

20000

40000

60000

80000

100000

120000

0 50 100 150 200

a[m/s2]19963,94s*k[mm] k=


x0=150 [mm]

hs =10 [mm]

hT=10 [mm]i=1; =12.7%rb=2 [mm]e=0 [mm]

legea: C4P1-5y=2x-x 2

Analiza dinamic la cama rotativ cu tachettranslant cu rol

a max =97000

s max =3.88

a min= -33000

Fig. 5. Law C4P1-5, Module B, u=80 [grad], n=40000 [rot/min]

-60000

-40000

-20000

0

2000040000

60000

80000

100000

0 50 100 150 200

a[m/s2]15044,81s*k[mm] k=


rb=3 [mm]b=30 [mm]d=30 [mm]x0=200 [mm]i=1;=16.5%

legea: C4P3-2y=2x-x 2

yc=1-x2

Analiza dinamic la cama rotativ cu tachetbalansier cu rol (Modul F) - A12

amax =80600

s max =4.28

a min= -40600

hT=15.70 [mm]

Fig. 6. Law C4P3-2, Module F, u=85 [grad], n=40000 [rot/min]


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3 Conclusions

Using the classical movement laws, the dynamics of the distribution cam-gears,spoils rapidly at the increasing of the rotation velocity of the shaft. To support a highrotation velocity is necessary the synthesis of the cam-profile by new movement laws,and at new Modules. A new and original movement law is presented in the picturesnumber 4, 5 and 6; it allows the increasing of the rotation velocity at the values: 10000-20000 [rot/min], at the classical module C presented (fig. 4). With others modules (B,F) one can obtains 30000-40000 [rot/min] (see the fig. 5, 6).

References

1- Petrescu F.I., Petrescu R.V., Contributions at the dynamics of cams . In the NinthIFToMM International Sympozium on Theory of Machines and Mechanisms,SYROM 2005, Bucharest, Romania, 2005, Vol. I, p. 123-128.

Relly Victoria PETRESCUUniversitatea Politehnica din Bucure ti, Departamentul GDGI

Splaiul Independen ei 313, Bucure ti, Sector 6, cod 060042 [email protected]

Adriana COM NESCUUniversitatea Politehnica din Bucure ti, Departamentul TMR Splaiul Independen ei 313, Bucure ti, Sector 6, cod 060042

[email protected]

Florian Ion PETRESCUUniversitatea Politehnica din Bucure ti, Departamentul TMR Splaiul Independen ei 313, Bucure ti, Sector 6, cod 060042

[email protected]

cam dynamics-02 tiberiu-petrescu

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